Duality Theory I: Basic Theory

Home , Bipolar theorem, Duality (mathematics)

DT I c Gabriel Nagy

Notes from the Functional Analysis Course (Fall 07 - Spring 08)

This section contains an important conceptual discussion on duality, which in a special case encodes the interplay between a locally convex space and its (topological) dual.

A. Dual Pairings

Convention. Throughout this note K will be one of the ﬁelds R or C, and all vector spaces are over K. Deﬁnitions. Suppose one has two vector spaces X and Y.A dual pairing of X with Y is a bilinear1 map Φ : X × Y → K, which is non-degenerate, in the following sense: • If x ∈ X is such that Φ(x, y) = 0, ∀ y ∈ Y, then x = 0.

• If y ∈ Y is such that Φ(x, y) = 0, ∀ x ∈ X , then y = 0. Equivalently, Φ can be represented by two2 linear injective maps

#: X 3 x 7−→ x# ∈ Y0, (1) #: Y 3 y 7−→ y# ∈ X 0, (2) satisfying the equality x#(y) = y#(x), ∀ x ∈ X , y ∈ Y. (3) The bilinear map Φ is simply given as Φ(x, y) = x#(y) = y#(x). Example 1. Given an arbitrary vector space X , and a linear subspace F ⊂ X 0, which separates the points of X (i.e. if x1 6= x2, there exists f ∈ F with f(x1) 6= f(x2)), then the map Φ : X × F 3 (x, f) 7−→ f(x) ∈ K is a dual pairing. Example 2. We can specialize the above example to the following situation. Assume X is a locally convex topological vector space. In particular the topology T, which X is equipped with, is Hausdorﬀ, and the dual space X ∗ = (X , T)∗ separates the points in X . Then the map Φ : X × X ∗ 3 (x, φ) 7−→ φ(x) ∈ K is a dual pairing. B. The Weak Dual Topologies

Deﬁnitions. Assume now a dual pairing is given, deﬁned by the maps (1) and (2) satisfying (3). If we endow Y0 with the weak* topology, we can apply the pull-back construction

1 This means that Φ is linear in each variable. 2 Although the notation is abusive, it is very convenient!

1 (see LCVS II) using the map (1) to obtain a topology, denoted by w# on X . Specifically, this is the weakest topology which makes the map #: X → (Y0, w∗) continuous. Since the weak* topology on Y0 is Hausdorff locally convex, and the map # is linear and injective, the topology w# on X is Hausdorff locally convex. Similarly, one also constructs a Hausdorff locally convex topology w# on Y, as the weakest topology that makes the map #: Y → (X 0, w∗) continuous. The two topologies constructed above are called the weak topologies associated with the dual pairing. Remarks 1-2. Use the notations as above. 1. Convergence in (X , w#) and in (Y, w#) can be characterized as follows:

w# (i) xλ −−→ x (in X ), if and only if one of the following equivalent conditions holds:

# # xλ (y) → x (y), ∀ y ∈ Y, # # y (xλ) → y (x), ∀ y ∈ Y.

w# (ii) yλ −−→ y (in X ), if and only if one of the following equivalent conditions holds:

# # yλ (x) → y (x), ∀ x ∈ X , # # x (yλ) → x (y), ∀ x ∈ X .

2. In terms of seminorms:

# X (i) the topology w on X is deﬁned by the family PY = {py}y∈Y , given by

# # py(x) = |y (x)| = |x (y)|, x ∈ X , y ∈ Y;

# Y (ii) the topology w on Y is deﬁned by the family PX = {px}x∈X , given by

# # px(y) = |x (y)| = |y (x)|, x ∈ X , y ∈ Y;

Theorem 1. Use the notations above.

(i) For any x ∈ X , the linear functional x# : Y → K is continuous in the w# topology. (i’) If we equip the space Y∗ = (Y, w#)∗ with the weak* topology, the map

#:(X , w#) 3 x 7−→ x# ∈ (Y∗, w∗) (4) is a topological linear isomorphism.

(ii) For any y ∈ Y, the linear functional y# : X → K is continuous in the w# topology. (ii’) If we equip the space X ∗ = (X , w#)∗ with the weak* topology, the map

#:(Y, w#) 3 y 7−→ y# ∈ (X ∗, w∗) (5) is a topological linear isomorphism.

2 Proof. By symmetry, it suffices to prove only (i) and (i’). Property (i) is quite trivial, by Remark 1. To prove (ii) we first remark that (4) is injective, by definition. Next we show that (4) is surjective. Suppose φ : Y → K is linear and w#-continuous, and let us prove that there exists (a unique) x ∈ X , such that φ = x#. Using Remark 2, the discussion from LCVS IV (Remark 3), there exist x1, . . . , xn ∈ X and t1, . . . , tn > 0, such that # # |φ(y)| ≤ t1|x1 (y)| + ··· + tn|xn (y)|, ∀ y ∈ Y. # Using Exercise ?? from HB, (applied to the seminorms pj(y) = tj|xj (y)|, j = 1, . . . , n), there exist linear functionals φ1, . . . , φn : Y → K, such that φ = φ1 + ··· + φn, and

# |φj(y)| ≤ tj|xj (y)|, ∀ y ∈ Y, j = 1, . . . , n.

# Fix for the moment j, and consider the linear functional ψj = tjxj : Y → K. Since |φj(y)| ≤ |ψj(y)|, ∀ y ∈ Y, we must have the inclusion Ker ψj ⊂ Ker φj, so X /Ker φj is a quotient of X /Ker ψj. Since ψj is linear, we must have

1 ≥ dim(Ker ψj) ≥ dim(Ker φj) ≥ 0,

so either

(i) dim(Ker ψj) = dim(Ker φj)(≤ 1), in which case Ker φj = Ker ψj, or

(i) 1 = dim(Ker ψj) > dim(Ker φj) = 0, in which case Ker φj = X .

In either case we get φj = αjψj, for some αj ∈ K (in case (ii), αj = 0). So now we have # φj = αjtjxj , j = 1, . . . , n, so

# # # φ(y) = α1t1x1 (y) + ··· + α1t1x1 (y) = (α1t1x1 + ··· + αntnxn) (y), ∀ y ∈ Y,

# i.e. φ = (α1t1x1 + ··· + αntnxn) . Having proven that (4) is a linear isomorphism, its continuity, as well as the continuity of ∗ # w # ∗ its inverse, are tautologic by Remark 1, since the condition xλ −→ x (in Y ) is equivalent to # # xλ (y) → x (y), ∀ y ∈ Y,

w# which in turn is equivalent to xλ −−→ x (in X ). We now re-visit the two examples from sub-section A. Example 1. Given an arbitrary vector space X , and a linear subspace F ⊂ X 0, which separates the points of X , then, as pointed out in sub-section A, we have is a dual pairing between X and X . The weak topology w# on X is referred to as the weak F-topology, and is denoted by wF . The other weak topology w# on F is simply the restriction of the weak* topology (from X 0) to F. Example 2. If (X , T) is a locally convex topological vector space (thus T is Hausdorﬀ), then, as discussed in Example 2 from sub-section A, we have a dual pairing between X and

3 ∗ # X . The topology w on X is simply the weak topology wT, implemented by T, which was introduced in LCVS IV. In particular (see LCVS IV), for every set S ⊂ X , one has the equalities T # conv S = conv SwT = conv Sw . (6) As before, the other topology w# on X ∗ is simply the weak* topology. As a consequence of Theorem 1, we also have a natural topological linear isomorphism

# ∗ ∗ ∗ #:(X , wT) 3 x 7−→ x ∈ (X , w ) , deﬁned by x#(φ) = φ(x), ∀ x ∈ X , φ ∈ X ∗. Exercise 1*. Equip the direct sum X = L with the locally convex sum topology N K T. Show that the weak topology wT is strictly weaker than T. Conclude that there does not exist locally convex spaces Y, such that (X , T) is topologically linearly isomorphic to (Y∗, w∗). Q Exercise 2. Let I be some non-empty set. Consider the spaces Xprod = K and L I Xsum = I K. 3 (i) Prove that the map Φ : Xprod × Xsum → K, deﬁned by X Φ(x, y) = xiyi, ∀ x = (xi)i∈I ∈ Xprod, y = (yi)i∈I ∈ Xsum, i∈I establishes a dual pairing.

# (ii) Prove that the topology w on Xprod coincides with the product topology Tprod.

# (iii) Prove that the topology w on Xsum is strictly weaker than the locally convex sum ∗ topology. Therefore we only have an algebraic linear isomorphism Xsum ' (Xprod, Tprod) .

C. Polars and the Bipolar Theorem

As we have already seen in Example 2, the closure of convex hulls depends only on the interaction between the ambient space and its (topological) dual. Therefore, it is expected that the operation of taking closed convex hulls to admit an “abstract” characterization, within the framework of dual pairs. Definitions. Suppose a dual pairing between two vector spaces X and Y is given, defined by the maps (1) and (2) satisfying (3). For a non-empty set A ⊂ X , we define its polar (in Y) to be the set: A◦ = {y ∈ Y : Re y#(a) ≤ 1, ∀ a ∈ A}. Similarly, for a non-empty set B ⊂ Y, we define its polar (in X ) to be the set:

B◦ = {x ∈ X : Re x#(b) ≤ 1, ∀ b ∈ B}.

Exercise 3. Prove that, using the notations as above, for two non-empty sets A ⊂ B ⊂ X , one has the reverse inclusion A◦ ⊃ B◦.

3 The sum is clearly ﬁnite, since yi 6= 0 only for ﬁnitely many i ∈ I.

4 Proposition 2. With the notations as above, if A ⊂ X is non-empty, then A◦ is convex, w#-closed in Y, and contains 0. Likewise, if B ⊂ Y is non-empty, then B◦ is convex, w#-closed in X , and contains 0. Proof. By symmetry, we only need to prove the ﬁrst statement. The fact that A◦ contains 0 is trivial. To prove convexity, start with y, z ∈ A◦ and some t ∈ [0, 1], and let us show that ty + (1 − t)z ∈ A◦. This is, however, trivial, since by the linearity of the map (5) we have Re[(ty + (1 − t)z)#(a)] = Re[ty#(a) + (1 − t)z#(a)] = = t[Re y#(a)] + (1 − t)[Re z#(a)] ≤ 1, ∀ a ∈ A.

◦ # ◦ # Finally, to prove that A is w -closed, we start with some net (yλ) in A , which is w - convergent to some y ∈ Y, and we show that y ∈ A◦. This is again trivial, since the # w # # # condition yλ −−→ y implies yλ (a) → y (a), ∀ a ∈ A, so the inequalities Re yλ (a) ≤ 1, ∀ λ, will force Re y#(a) ≤ 1. Theorem 2 (Bipolar Theorem). Use the notations and hypotheses from the above deﬁ- nition. Let w# be the weak topology on X associated with the dual pairing. For any subset A ⊂ X , the w#-closure of the convex hull of A ∪ {0} is given by the equality

w# conv(A ∪ {0}) = (A◦)◦. (7) Proof. Using Proposition 2, we already know that the bi-polar (A◦)◦ is w#-closed (in X ), convex, and contains 0. Note also that (A◦)◦ contains A. Indeed, if a ∈ A, then for every y ∈ A◦ one has a#(y) = y#(a), so we now have Re a#(y) = Re y#(a) ≤ 1, ∀ y ∈ A◦. (The second inequality comes from the very deﬁnition of A◦.) Since (A◦)◦ is w#-closed, convex, and it contains A ∪ {0}, we have the inclusion

w# conv(A ∪ {0}) ⊂ (A◦)◦. To prove the other inclusion, we start with some element x ∈ X that does not belong to w# conv(A ∪ {0}) , and we show that x 6∈ (A◦)◦. First of all, by Theorem 1 from LCVS IV, we know there exist φ ∈ (X , w#)∗ and s ∈ R, such that Re φ(a) ≤ s < Re φ(x), ∀ a ∈ A ∪ {0}. In particular (using a = 0), it follows that Re φ(x) > s ≥ 0, so if we take t = (Re φ(x)+s)/2, and we define ψ = t−1φ, then ψ : X → K is still linear and w#-continuous, but now satisfies Re ψ(a) < 1 < Re ψ(x), ∀ a ∈ A ∪ {0}. By Theorem 1, we know that there exists (a unique) y ∈ Y, such that ψ = y#, so the above inequalities now read: Re y#(a) < 1 < Re y#(x), ∀ a ∈ A ∪ {0}. In particular, from the first inequality we can conclude that y ∈ A◦, and then the second inequality shows that x 6∈ (A◦)◦.

5 Comment. Many textbooks use a notion that is slightly diﬀerent from ours, by deﬁning (use the notations as above), for non-empty A ⊂ X and B ⊂ Y, the sets

# A = {y ∈ Y : |y (a)| ≤ 1, ∀ a ∈ A}, # B = {x ∈ X : |x (b)| ≤ 1, ∀ b ∈ B}.

Deﬁnition. With the above notations, the set A is called the absolute polar of A (in Y), and the set B is called the absolute polar of B (in X ). The result below deal with the “absolute” version of Proposition 2 and the relationship between the absolute and the “honest” polars. Proposition 3 Use the notations as above. If A ⊂ X is a non-empty subset, then:

# (i) The absolute polar A is w -closed, convex, and balanced (hence A 3 0);

◦ (ii) A = [bal A] . S (Recall that bal A denotes the balanced hull of A, deﬁned as α∈K αA.) |α|≤1

◦ # Proof. (ii). Suppose ﬁrst y ∈ A, and let us prove that y ∈ [bal A] , i.e. Re y (αa), for all a ∈ A, and every α ∈ K with |α| ≤ 1. This is, however, obvious, since Re y#(αa) ≤ |y#(αa)| = |α| · |y#(a)| ≤ |α| ≤ 1.

◦ Conversely, if y ∈ [bal A] , then we can choose, for every a ∈ A, a scalar αa ∈ K such that # # |αa| = 1 and αay (a) = |y (a)|(∈ R). In particular, for every a ∈ A, we have

# # # # |y (a)| = αay (a) = Re[αay (a)] = Re y (αaa) ≤ 1

(the last inequality follows from the assumption that y is in [bal A]◦), so y indeed belongs to A. # (i). By part (ii) and Proposition 2, the absolute polar A is w -closed, convex (and contains 0). If α ∈ K is such that |α| ≤ 1, then for every y ∈ A we have: |(αy)#(a)| = |α| · |y#(a)| ≤ |α| ≤ 1, which means that αy is also in A. The “absolute” version of Theorem 2 is the following. Theorem 3 (Absolute Bipolar Theorem) Use the notations and hypotheses from the preceding deﬁnition. Let w# be the weak topology on X associated with the dual pairing. For any non-empty subset A ⊂ X , one has the equality

w# conv(bal A) = (A). (8)

6 Proof. By Proposition 3, we know that

◦ A = [bal A] . (9)

◦ Since A is already balanced, again by Proposition 3, we get the equality (A) = (A) , so going back to (9) we now have ◦ ◦ (A) = ([bal A] ) . The desired equality (8) now follows from the one above, and Theorem 2 (applied to bal A).

Exercise 4. Use the notations as above. Prove that, for a non-empty set A ⊂ X , one has: (i) (tA)◦ = t−1(A◦), ∀ t > 0;

−1 (ii) (αA) = |α| (A, ∀ α ∈ K, α 6= 0. Besides (absolute) polars, there is a third construction associated with dual pairs, de- scribed in the following group of Exercises. Exercises 5-7. Use notations as above. For non-empty A ⊂ X and B ⊂ Y, let: Aann = {y ∈ Y : y#(a) = 0, ∀ a ∈ A}, Bann = {x ∈ X : x#(b)y#(a) = 0, ∀ b ∈ B}. 5. Prove the equalities4

ann ◦ ann A = (span A) = (span A) = (span A).

6. Prove that Aann is a w#-closed linear subspace in Y.

w# 7. Prove that (Aann)ann = span A . Deﬁnition. The set Aann is called5 the annihilator of A (in Y), and the set Bann is called the annihilator of B (in X ). Comment. The (Absolute) Bipolar Theorem has numerous application in Functional Analysis, especially in the setting from Example 2. Speciﬁcally, if (X , T) is a locally convex topological vector space, we can from a dual pairing with its topological dual X ∗ = (X , T)∗, and then by Theorems 3 and 4, it follows that, for every A ⊂ X , one has the equalities

T conv(A ∪ {0}) = (A◦)◦; (10) T conv(bal A) = (A). (11) Some subtle applications are illustrated in DT II.

Exercise 8. Let op be any one of the three operations: “◦” (polar), “” (absolute polar), or “ann” (annihilator). Prove that, given a dual pairing between X and Y, for a non-empty set A ⊂ X one has the equality [(Aop)op]op = Aop.

4 Here span A denotes the linear span of A. 5 Some texts refer to this as the orthogonal of A in Y, and denote it by A⊥.